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Show that the $5$-sylow subgroup of $S_{24}$ is isomorphic to $\mathbb{Z}_5\ \times \mathbb{Z}_5 \times \mathbb{Z} _5 \times \mathbb{Z}_5$.

So as I understand this, I need to prove that every element in the $5$-sylow subgroup is a multiplication of $4$ cycles of length $5$, and the rest are cycle of length $1$.

Which makes sense, as the elements should be of order $5$, and cycles of length $5$ are of order $5$, so I can see why it should be isomorphic to $\mathbb{Z}_5\ \times \mathbb{Z}_5 \times \mathbb{Z} _5 \times \mathbb{Z}_5$, as $4$ cycles of length $5$, and another $4$ elements of order $1$ (since it is $S_{24}$).

How can I prove it formally?(Assuming all I said is correct... Otherwise, any hints?)

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    $\begingroup$ What is the order of the Sylow subgroups of $S_{24}$? Can you explicitly write down a subgroup of that order? $\endgroup$ – Alex B. Jun 23 '18 at 12:51
  • $\begingroup$ Can you write down four "independent" commuting elements of order $5$? $\endgroup$ – Angina Seng Jun 23 '18 at 12:51
  • $\begingroup$ @AlexB. The order would be the $p$ prime for each $p$-sylow sub group.... I believe that the order of $S_{24}$ is $24!$? Therefore it would be a bit difficult to find all prime number for all possible sylow subgroup... $\endgroup$ – ChikChak Jun 23 '18 at 13:01
  • $\begingroup$ @LordSharktheUnknown What do you mean by "independent"? $\endgroup$ – ChikChak Jun 23 '18 at 13:03
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A $5$-Sylow subgroup of $S_{24}$ has order $5^4$ because $24!=1 \cdots 5 \cdots 10\cdots 15 \cdots 20 \cdots 24$ has four $5$ factors.

The $5$-cycles $(1,2,3,4,5)$, $(6,7,8,9,10)$, $(11,12,13,14,15)$, $(16,17, 18,19,20)$ commute and so generate a subgroup of order $5^4$ isomorphic to $\mathbb{Z}_5 \times \mathbb{Z}_5 \times \mathbb{Z} _5 \times \mathbb{Z}_5$.

This subgroup must be a $5$-Sylow subgroup of $S_{24}$.

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  • $\begingroup$ Every $p$-sylow subgroup of $S_n$ is always calculated that way? Could you elaborate why those cycles generate a group of that order? $\endgroup$ – ChikChak Jun 23 '18 at 13:35
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    $\begingroup$ @ChikChak, see en.wikipedia.org/wiki/Symmetric_group#Sylow_subgroups $\endgroup$ – lhf Jun 23 '18 at 13:36

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